MATH 510501 Probability and Statistics I; Fall 2003
Homework 6; Problem 3.5.6
Tony Smaldone
Problem 3.5.6
Suppose that in a certain drug the concentration of a particular chemical is a random
variable with a continuous distribution for which the p.d.f. is as follows:
g
(
x
) =
3
8
x
2
for 0
≤
x
≤
2
,
0 otherwise
.
Suppose that the concentrations
X
and
Y
of the chemical in two separate batches of the drug are independent
random variables for each of which the p.d.f. is
g
. Determine
a. the joint p.d.f. of
X
and
Y
b.
P
(
X
=
Y
)
c.
P
(
X > Y
)
d.
P
(
X
+
Y
≤
1)
Solution:
a. Given that
X
and
Y
are independent, the following relationship is true
f
(
x, y
) =
f
X
(
x
)
f
Y
(
y
)
.
Since
X
and
Y
both have the p.d.f.
g
, then
f
(
x, y
) =
g
(
x
)
g
(
y
). Hence
f
(
x, y
) =
3
8
x
2 3
8
y
2
=
9
64
x
2
y
2
,
0
≤
x
≤
2 and 0
≤
y
≤
2
,
0
,
otherwise
.
b. To evaluate
P
(
X
=
Y
) we need to define the region of integration. Here we have
y
=
x
, so the limits
for y are
x
≤
y
≤
x
and 0
≤
x
≤
2 for x. Hence,
P
(
X
=
Y
) =
2
0
x
x
9
64
x
2
y
2
dydx
= 0
.
As discussed in problem 3.4.4, the probability that
X
and
Y
will lie on a straight line is zero.
c. To evaluate
P
(
X > Y
) we need to define the region of integration. Here we have
y < x
, so the limits
for y are 0
≤
y
≤
x
and are 0
≤
x
≤
2 for x. Hence,
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 Spring '08
 Staff
 Statistics, Probability, Probability distribution, Probability theory, Tony Smaldone, inner integral yields

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